An electron is moved from point A to point B in a uniform electric field and gains 4.66×10-15 J of electrostatic potential energy. Calculate the magnitude of the electrostatic potential difference between the two points. (in V)

2.9125 x 10⁴ V

Explanation:

The electrostatic potential difference (ΔV) between two points, say A and B, is denoted by - and is the quotient of the change in the electrostatic potential energy (Δ) of a charge Q moved from A to B, and the charge. i.e

ΔV = - = Δ / Q               ------------(i)

From the question;

Δ = 4.66 x 10⁻¹⁵J

Q = charge of an electron = -1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

- = (4.66 x 10⁻¹⁵) / (-1.6 x 10⁻¹⁹)

- = -2.9125 x 10⁴ V

| - | = 2.9125 x 10⁴ V

Therefore, the magnitude of the electrostatic potential difference between these two points is 2.9125 x 10⁴ V

a electromagnetic waves

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